How much electricity in terms of Faraday is required to produce -
40.0 g of Al from molten `Al_(2)O_(3)` ?

Explanation

The electrode reaction is : `Al^(3+)+3e rarr Al`
Thus 1 mol of `Al(27 g)` requires 3F electricity
`therefore` 40 g of Al requires `=(3)/(27)xx40=4.44F`